Replace a word with multiple lines using sed?

Question

I\'m working on a bash-script that has to prepare an E-Mail for being sent to a user.

It aggregates some data, which ends up being multiple lines of stuff. For the e

Answer 1

Echo variable into temporary text file.

Insert text file into mail.tpl and delete data from mail.tpl

echo ${DATA} > temp.txt    
sed -i -e "/_data_/r temp.txt" -e "//d" mail.tpl

Answer 2

If you build your multiple line text with "\n"s, this will work with a simple sed command as:

DATA=`echo ${DATA} | tr '\n' "\\n"`
#now, DATA="line1\nline2\nline3"
sed "s/_data_/${DATA}/" mail.tpl

Answer 3

Not sure if you have tried to put "\n" in the replace part

sed 's/[pattern]/\
[line 1]\n\
[line 2]\n\
[line n]\n\
/g' mail.tpl

The first line has /\ for readibility reasons. Each line after that is a stand-alone line like you would find in a text editor. Last line is stand-alone, once again for readability reasons. You can make all of this one line if needed. Works on Debian Jessie when I tested it.

Answer 4

You can put your data in a temp file and run:

$ sed '/_data_/r DATA_FILE' mail.tpl | sed '/_data_/d'> temp; mv temp mail.tpl

Answer 5

I tried it and sed 's/pattern/\na\nb\nc/g' but it does not work on all systems. What does work is putting a \ followed by a newline in the replace pattern, like this:

sed 's/pattern/a\
b\
c/g'

This appends a line containing b and a line containing c when the pattern is seen.

To put it in a variable, use double backslashes:

export DATA="\\
a\\
b\\
c"

and then:

sed "s/pattern/${DATA}/g"

Note the double quotes.

Answer 6

Escaping all the newlines with a \ (except the last one) worked for me. The last newline must not be escaped not to break the s command.

Example :

DATA="a
b
c"

ESCAPED=$(echo "${DATA}" | sed '$!s@$@\\@g')
echo "${ESCAPED}" 
a\
b\
c

sed "s/pattern/${ESCAPED}/" file